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Oct 3 Square root of a square root

MathJax TeX Test Page$$\sqrt[4]{161 - 72\sqrt(5)}$$
Our goal is to completely eliminate the root. Initially, it seems very difficult to solve. An important thing about these problems is that the expression inside of the fourth root is a square. So $\left(A\sqrt{5} + B\right)^2 = 161 - 72\sqrt{5}$
So, we expand the left side:
$$5a^2 + 2ab\sqrt{5} + b^2 = 161 - 72\sqrt{5}$$
We can turn these into two equations:
$$5a^2 + b^2 = 161$$$$2ab\sqrt{5} = -72\sqrt{5}$$
Looking at the equations together and doing a bit of guess and check, we get $(a,b) = (\pm4, \mp9)$.
We have to make sure the expression is positive however, because our final answer is
$$\sqrt{\pm4\sqrt{5} \mp9}$$
$4\sqrt{5} = \sqrt{80} < 9$, so it must be $9 - 4\sqrt{5}$. So, our answer is:
$$\sqrt{9 - 4\sqrt{5}}$$
Now, we say that $9 - 4\sqrt{5}$ is a square itself, so
$$9 - 4\sqrt{5} = \left(c\sqrt{5} + d\right)^2$$
$$9 - 4\sqrt{5} = 5c^2 + d^2 + 2dc\sqrt{5}$$
Once again we get these two equations:
$$9 = 5c^2 + d^2$$
$$-2 = dc$$
Our solution is $(c,d) = (\pm1, \mp2)$, which can be found with guess and check. $\sqrt{5} > 2$, so our final answer is:
$$\boxed{\sqrt[4]{161 - 72\sqrt(5)} = \sqrt{5} - 2}$$

Conclusion

So, you kinda have to do some guess and check to get it right. This type of question will definitely come up in a math competition, and the numbers will usually be pretty nice, so this is a useful skill to know.